# Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 13)
13.
You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
172.16.17.1 255.255.255.252
172.16.0.1 255.255.240.0
172.16.20.1 255.255.254.0
172.16.18.255 255.255.252.0
Explanation:
A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
Discussion:
13 comments Page 1 of 2.

Taran said:   7 years ago
Hi Guys!

We have first 6 on-bits so after calculating the values.
128 + 64 + 32 + 16 + 8 + 4 = 252, we get sum of on-bit value.

Block size = 256 - Sum of on bit value.

256 - 252 = 4,
The block size is 4.

Subnets in the network
172.16.0.0/22
172.16.4.0/22
172.16.8.0/22
172.16.12.0/22
172.16.16.0/22
172.16.20.0/22 -------- and so on up to 172.16.252.0

So to here some of you confused that why A is not correct the reason behind it is
that if you select A, as you answer then you are Host bits are showing as network bits bcoz we have an ip of B class and in B class we deal with N.N.H.H, so we are not selecting A as the correct answer due to the wrong Subnet.

D is dealing with class B Subnet.
(172. 16 . 18 . 255)
(255. 255. 252. 0 )
( N . N . H . H )
(1)

Taran said:   7 years ago
@Reddy.

D is the right answer bcoz of it's subnet.
class B subnet is always N:N:H:H bits.
172.16.18.255 is also in between the correct range.
A option comes under the C Class subnet tata's why we choose D as the correct answer.

Monty said:   6 years ago
According to the question, 172.16.17.0/22 this is given but while solving u said in question 172.16.16.0 is used which makes only option D as a correct answer but as per real question, both A & D can be right.

Gio said:   3 years ago
Let me explain.
We learned that subnets with CIDR /22 can skip at 4 size block.

So, how is possible that the question subnet is 17.0, when the only ones available are 16.0, 20.0 etc?

(1)

Pruthvi said:   10 years ago
11111111.11111111.1 1 1 1 1 {1} 0 0.00000000/22

128 64 32 16 8 {4} 2 1

Here the host and network parts are separated at 4. So the block size is 4. This is a simple trick.

Usha said:   5 years ago
As per question, we need valid host address and the correct option is D, can you explain how 172.16.18.255 is valid host address?

Haya said:   8 years ago
Why option A is not correct?

It is also a valid host. Because range is in between 172.16.16.1 to 172.16.19.254.

Sechonge said:   8 years ago
We know the bi borrowed id 6, ie 11111100 = 252 and we deal with N.N.H.H.

That's why the answer is D.

Nitz said:   8 years ago
Hi, I can not understand how they got it 172.16.18.255, Can anyone explain me in detail?
(1)

Mtk said:   10 years ago
I couldn't understand how did the /22 mask could have block size of 4?